Isotropy irreducible riemannian manifolds have an unique $G$-invariant metric

Question

Let $G/H$ be a quotient of Lie group $G$ on a closed subgroup $H$. And let $$\text{is}:H\to GL(T_0(G/H))$$ be the isotropy representation of $H$. I managed to prove the following statement (that constitutes the first part of the exercise):

If $\langle -,- \rangle $ is an inner product on $T_0(G/H)$ and $A$ is an endomorphism of $T_0(G/H)$, we can define a new inner product: $$\langle\langle-,-\rangle \rangle :=\langle A-,-\rangle.$$ Prove that if $\langle -,- \rangle$ and $\langle \langle -,- \rangle \rangle$ are both $\text{is}(H)$-invariant, then the elements of $\text{is}(H)$ preserve the eigenspaces of $A$.

Now I have to prove (using the first part of the exercise) that if the isotropy representation of $H$ is irreducible, then there is an unique $G$-invariant metric on $G/H$ (up to a scalar). First of all, I have a question on the terminology:

• When we say "$G$-invariant" we mean that it's invariant with respect to the action of $G$ on $G/H$ from the left or from the right (or both)?

And second thing, I tried to prove this but I didn't manage to. I know that there are similar questions on this site, but they almost all use Schur's Lemma, while I should use the first part of this exercise. I'll present my attempt:

My attempt Let $g^1$ and $g^2$ be two $G$-invariant metrics. Since they are invariant we can restrict to working on $T_0(G/H)$ and if we fix a non-zero vector $v\in T_0(G/H)$, I can assume that $g^1(v,v)=g^2(v,v)$ (by eventually scaling one of them). Now my idea was to define something like: $$T:=\{u\in T_0(G/H): g^1(v,u)=g^2(v,u)\}$$ (this is just a sketch, I had also some other possible definitions in mind). And I wanted to prove that $T$ or something like that was a submodule of the representation to use the irreducibility hyphothesis.

Answer 1

First, $G$ acts on $G/H$ only from the left, so invariance is only meant with respect to that action.

Hint for solving the exercise: Linear algebra tells you that there is a linear map $A:T_o(G/H)\to T_o(G/H)$ such that $g^2(v,w)=g^1(Av,w)$ for $v,w\in T_o(G/H)$. Moreover, $A$ is symmetric with respect to $g^1$ and hence diagonalizable. Then use the part you solved already and irreducibility.